Non-Relativistic Propagators Via Schwinger's Method

1260 Brazilian Journal of Physics, vol. 37, no. 4, December, 2007

Non-Relativistic Propagators via Schwinger’s Method

A. Aragao,˜ H. Boschi-Filho, C. Farina, Instituto de F´ısica, Universidade Federal do Rio de Janeiro, Caixa Postal 68.528, 21941-972 Rio de Janeiro, RJ, Brazil.

and F. A. Barone Universidade Federal de Itajuba,´ Av. BPS 1303 Caixa Postal 50 - 37500-903, Itajuba,´ MG, Brazil.

Received on 10 September, 2007 In order to popularize the so called Schwinger’s method we reconsider the Feynman propagator of two non- relativistic systems: a charged particle in a uniform magnetic field and a charged harmonic oscillator in a uniform magnetic field. Instead of solving the Heisenberg equations for the position and the canonical momentum operators, R and P, we apply this method by solving the Heisenberg equations for the gauge invariant operators R and π = P − eA, the latter being the mechanical momentum operator. In our procedure we avoid fixing the gauge from the beginning and the result thus obtained shows explicitly the gauge dependence of the Feynman propagator. Keywords: Schwinger’s method, Feynman Propagator, Magnetic Field, Harmonic Oscillator.

I. INTRODUCTION knowledge, since then only a few papers have been written with this method, namely: in 1986, Urrutia and Manterola [18] used it in the problem of an anharmonic charged oscil- In a recent paper [1], three methods were used to com- lator under a magnetic field; in the same year, Horing, Cui, pute the Feynman propagators of a one-dimensional harmonic and Fiorenza [19] applied Schwinger’s method to obtain the oscillator, with the purpose of allowing a student to com- Green function for crossed time-dependent electric and mag- pare the advantadges and disadvantadges of each method. netic fields; the method was later applied in a rederivation The above mentioned methods were the following: the so of the Feynman propagator for a harmonic oscillator with a called Schwinger’s method (SM), the algebraic method and time-dependent frequency [20]; a connection with the mid- the path integral one. Though extremely powerful and ele- point-rule for path integrals involving electromagnetic inter- gant, Schwinger’s method is by far the less popular among actions was discussed in [21]. Finally, pedagogical presen- them. The main purpose of the present paper is to popularize tations of this method can be found in the recent publication Schwinger’s method providing the reader with two examples [1] as well as in Schwinger’s original lecture notes recently slightly more difficult than the harmonic oscillator case and published [22], which includes a discussion of the quantum whose solutions may serve as a preparation for attacking rela- action principle and a derivation of the method to calculate tivistic problems. In some sense, this paper is complementary propagators with some examples. to reference [1]. The method we shall be concerned with was introduced by It is worth mentioning that this same method was inde- Schwinger in 1951 [2] in a paper about QED entitled “Gauge pendently developed by M. Goldberger and M. GellMann in invariance and vacuum polarization”. After introducing the the autumn of 1951 in connection with an unpublished paper proper time representation for computing effetive actions in about density matrix in statistical mechanics [23]. QED, Schwinger was faced with a kind of non-relativistic Our purpose in this paper is to provide the reader with two propagator in one extra dimension. The way he solved this other examples of non-relativistic quantum propagators that problem is what we mean by Schwinger’s method for com- can be computed in a straightforward way by Schwinger’s puting quantum propagators. For relativistic Green functions method, namely: the propagator for a charged particle in a of charged particles under external electromagnetic fields, the uniform magnetic field and this same problem with an addi- main steps of this method are summarized in Itzykson and Zu- tional harmonic oscillator potential. Though these problems ber’s textbook [3] (apart, of course, from Schwinger’s work have already been treated in the context of the quantum action [2]). Since then, this method has been used mainly in rela- principle [18], we decided to reconsider them for the follow- tivistic quantum theory [4–16]. ing reasons: instead of solving the Heisenberg equations for However, as mentioned before, Schwinger’s method is also the position and the canonical momentum operators, R and well suited for computing non-relativistic propagators, though P, as is done in [18], we apply Schwinger’s method by solv- it has rarely been used in this context. As far as we know, this ing the Heisenberg equations for the gauge invariant opera- method was used for the first time in non-relativistic quantum tors R and π = P − eA, the latter being the mechanical mo- mechanics by Urrutia and Hernandez [17]. These authors used mentum operator. This is precisely the procedure followed by Schwinger’s action principle to obtain the Feynman propaga- Schwinger in his seminal paper of gauge invariance and vac- tor for a damped harmonic oscillator with a time-dependent uum polarization [2]. This procedures has some nice proper- frequency under a time-dependent external force. Up to our ties. For instance, we are not obligued to choose a particular A. Aragao˜ et al. 1261 gauge at the beginning of calculations. As a consequence, (i) we solve the Heisenberg equations for X(τ) and P(τ), and we end up with an expression for the propagator written in write the solution for P(0) only in terms of the operators an arbitrary gauge. As a bonus, the transformation law for X(τ) and X(0); the propagator under gauge transformations can be readly obtained. (ii) then, we substitute the results obtained in (i) into the ex- In order to prepare the students to attack more complex pression for H (X(0),P(0)) in (3) and using the com- problems, we solve the Heisenberg equations in matrix form, mutator [X(0),X(τ)] we rewrite each term of H in a which is well suited for generalizations involving Green func- time ordered form with all operators X(τ) to the left tions of relativistic charged particles under the influence of and all operators X(0) to the right; electromagnetic fields (constant Fµν, a plane wave field or (iii) with such an ordered hamiltonian, equation (3) can be even combinations of both). For pedagogical reasons, at the readly cast into the form end of each calculation, we show how to extract the corresponding energy spectrum from the Feynman propagator. Al- ∂ i~ K(x,x0;τ) = F(x,x0;τ)K(x,x0;τ), (5) though the way Schwniger’s method must be applied to non- ∂τ relativistic problems has already been explained in the litera- with F(x,x0;τ) being an ordinary function defined as ture [1, 18, 22], it is not of common knowledge so that we start this paper by summarizing its main steps. The paper is orga- hx,τ|H (X(τ),X(0))|x0,0i F(x,x0;τ) = ord . (6) nized as follows: in the next section we review Schwinger’s hx,τ|x0,0i method, in section III we present our examples and section IV is left for the final remarks. Integrating in τ, the Feynman propagator takes the form ½ Z ¾ i τ K(x,x0;τ) = C(x,x0)exp − F(x,x0;τ0)dτ0 , (7) II. MAIN STEPS OF SCHWINGER’S METHOD ~ 0 whereRC(x,x ) is an integration constant independent of For simplicity, consider a one-dimensional time- τ and τ means an indefinite integral; independent Hamiltonian H and the corresponding non- 0 relativistic Feynman propagator defined as (iv) last step is concerned with the evaluation of C(x,x ). This is done after imposing the following conditions h−iH τi K(x,x0;τ) = θ(τ)hx|exp |x0i, (1) ∂ ~ −i~ hx,τ|x0,0i = hx,τ|P(τ)|x0,0i, (8) ∂x 0 ∂ where θ(τ) is the Heaviside step function and |xi, |x i are the i~ hx,τ|x0,0i = hx,τ|P(0)|x0,0i, (9) eingenkets of the position operator X (in the Schrodinger¨ pic- ∂x0 0 ture) with eingenvalues x and x , respectively. The extension as well as the initial condition for 3D systems is straightforward and will be done in the next section. For τ > 0 we have, from equation (1), that lim K(x,x0;τ) = δ(x − x0) . (10) τ→0+ ∂ h−iH τi i~ K(x,x0;τ) = hx|H exp |x0i. (2) Imposing conditions (8) and (9) means to substitute in their ∂τ ~ left hand sides the expression for hx,τ|x0,0i given by (7), while Inserting the unity 1l = exp[−(i/~)H τ]exp[(i/~)H τ] in the in their right hand sides the operators P(τ) and P(0), respec- r.h.s. of the above expression and using the well known re- tively, written in terms of the operators X(τ) and X(0) with lation between operators in the Heisenberg and Schrodinger¨ the appropriate time ordering. pictures, we get the equation for the Feynman propagator in the Heisenberg picture, III. EXAMPLES ∂ i~ K(x,x0;τ) = hx,τ|H (X(0),P(0))|x0,0i, (3) ∂τ A. Charged particle in an uniform magnetic field where |x,τi and |x0,0i are the eingenvectors of operators X(τ) As our first example, we consider the propagator of a non- and X(0), respectively, with the corresponding eingenvalues relativistic particle with electric charge e and mass m, submit- x and x0: X(τ)|x,τi = x|x,τi and X(0)|x0,0i = x0|x0,0i, with ted to a constant and uniform magnetic field B. Even though K(x,x0;τ) = hx,τ|x0,0i. Besides, X(τ) and P(τ) satisfy the this is a genuine three-dimensional problem, the extension of Heisenberg equations, the results reviewed in the last section to this case is straightforward. Since there is no electric field present, the hamil- dX dP i~ (τ) = [X(τ),H ] ; i~ (τ) = [P(τ),H ]. (4) tonian can be written as dτ dτ (P − eA)2 π2 H = = , (11) Schwinger’s method consists in the following steps: 2m 2m 1262 Brazilian Journal of Physics, vol. 37, no. 4, December, 2007 where P is the canonical momentum operator, A is the vector Integrating equation (19) we find potential and π = P − eA is the gauge invariant mechanical Π(τ) = e2ωCτΠ(0) . (21) momentum operation. We choose the axis such that the magnetic field is given by B = Be3. Hence, the hamiltonian can be Substituting this solution in equation (18) and integrating once decomposed as more, we get

π2 + π2 P2 P2 sin(ωτ) ωCτ H = 1 2 + 3 = H + 3 , (12) R(τ) − R(0) = e Π(0) , (22) 2m 2m ⊥ 2m mω where we used the following properties of C matrix: with an obvious definition for H . ⊥ C2 = −1l; C−1 = −C = CT , eαC = cos(α)1l + sin(α)C with Since the motion along the OX direction is free, the three- 3 CT being the transpose of C. Combining equations (22) and dimensional propagator K(x,x0;τ) can be written as a prod- (21) we can write Π(0) in terms of the operators R(τ) and uct of a two-dimensional propagator, K (r,r0;τ), related to ⊥ R(0) as the magnetic field and a one-dimensional free propagator, µ ¶ (0) 0 mω K3 (x3,x3;τ): Π(0) = e−ωCτ R(τ) − R(0) . (23) sin(ωτ) 0 0 (0) 0 K(x,x ;τ) = K⊥(r,r ;τ)K3 (x3,x3;τ), (τ > 0) (13) 2 2 In order to express H⊥ = (π1 + π2)/2m in terms of R(τ) and (0) 0 R(0), we use (23). In matrix notation, we have where r = x1e1 + x2e2 and K3 (x3,x3;τ) is the well known propagator of the free particle [24], 1 H = ΠT (0)Π(0) r ⊥ 2m m h im (x − x0 )2 i µ K(0)(x ,x0 ;τ) = exp 3 3 . (14) mω2 3 3 3 2πi~τ 2~ τ = RT (τ)R(τ) + RT (0)R(0) + 2sin2 (ωτ) In order to use Schwinger’s method to compute the two- ¶ 0 0 T T dimensional propagator K⊥(r,r ;τ) = hr,τ|r ,0i, we start by −R (τ)R(0) − R (0)R(τ) . (24) writing the differential equation Last term on the r.h.s. of (24) is not ordered appropriately as ∂ 0 0 required in the step (ii). The correct ordering may be obtained i~ hr,τ|r ,0i = hr,τ|H⊥(R⊥(0),π⊥(0))|r ,0i , (15) ∂τ as follows: first, we write where R⊥(τ) = X1(τ)e1 + X2(τ)e2 and π⊥(τ) = π1(τ)e1 + 2 0 T T π2(τ)e2. In (15) |r,τi and |r ,0i are the eigenvectors of R(0) R(τ) = R(τ) R(0) + ∑[Xi(0),Xi(τ)]. (25) i=1 position operators R(τ) = X1(τ)e1 + X2(τ)e2 and R(0) = X1(0)e1 +X2(0)e2, respectively. More especifically, operators Using equation (22), the usual commutator [Xi(0),π j(0)] = 0 0 X1(0), X1(τ), X2(0) and X2(τ) have the eigenvalues x1, x1, x2 i~δi j1l and the properties of matrix C it is easy to show that and x2, respectively. 2 In order to solve the Heisenberg equations for operators R⊥(τ) 2i~sin(ωτ)cos(ωτ) ∑[Xi(0),Xi(τ)] = , (26) and π⊥(τ), we need the commutators i=1 mω h i 2 so that hamiltonian H with the appropriate time ordering Xi(τ),π j (τ) = 2i~πi(τ), ⊥ h i takes the form 2 ½ ¾ πi(τ),π j (τ) = 2i~eBεi j3π j(τ), (16) 2 mω 2 2 T H⊥ = 2 R (τ) + R (0) − 2R (τ)R(0) where εi j3 is the usual Levi-Civita symbol. Introducing the 2sin (ωτ) matrix notation − i~ωcot(ωτ). (27) µ ¶ µ ¶ X (τ) π (τ) Substituting this hamiltonian into equation (15) and integrat- R(τ) = 1 ; Π(τ) = 1 , (17) X2(τ) π2(τ) ing in τ, we obtain ½ ¾ C(r,r0) imω and using the previous commutators the Heisenberg equations hr,τ|r0,0i = exp cot(ωτ)(r − r0)2 , (28) of motion can be cast into the form sin(ωτ) 2~ dR(τ) Π(τ) where C(r,r 0) is an integration constant to be determined by = , (18) dτ m conditions (8), (9) and (10), which for the case of hand read dΠ(τ) µ ¶ = 2ωCΠ(τ), (19) 0 ∂ 0 dτ hr,τ|π j(τ)|r ,0i = −i~ − eA j(r) hr,τ|r ,0i (29) ∂x j Ã ! where 2ω = eB/m is the cyclotron frequency and we defined 0 ∂ 0 0 the anti-diagonal matrix hr,τ|π j(0)|r ,0i = i~ 0 − eA j(r ) hr,τ|r ,0i , (30) µ ¶ ∂x j 0 1 C = . (20) lim hr,τ|r0,0i = δ(2)(r − r0). (31) −1 0 τ→0+ A. Aragao˜ et al. 1263

In order to compute the matrix element on the l.h.s. of (29), equation (38) can be written as ½ ¾ we need to express Π(τ) in terms of R(τ) and R(0). From Z r 0 ie £ 1 ¡ 0¢¤ equaitons (21) and (23), we have C(r,r )= C0 exp A(ξ) − B × ξ − r ·dξ . ~ Γ 0 2 µ ¶ r mω (41) Π(τ) = eωτC R(τ) − R(0) , (32) sin(ωτ) Observe, now, that the integrand in the previous equation has a vanishing curl, which leads to the matrix element · ¸ 1 ¡ 0¢ ¡ ¢ ∇ × A(ξ) − B × ξ − r = B − B = 0 , 0 0 ξ 2 hr,τ|π j(τ)|r ,0i = mω[cot(ωτ) x j − x j ¡ ¢ 0 0 which means that the line integral in (41) is path independent. + ε jk3 xk − xk ]hr,τ|r ,0i , (33) Choosing, for convenience, the straightline from r 0 to r, it can where we used the properties of matrix C and Einstein con- be readly shown that vention for repeated indices is summed. Analogously, the Z r ¡ 0¢ l.h.s. of equation (30) can be computed from (23), Γ [B × ξ − r ] · dξ = 0 , sl r 0 ¡ ¢ 0 0 0 hr,τ|π j(0)|r ,0i = mω[cot(ωτ) x j − x j where Γ means a straightline from r to r. With this simpli- ¡ ¢ sl 0 0 fication, the C(r 0,r) takes the form − ε jk3 xk − x ]hr,τ|r ,0i . (34) k ½ ¾ Z r 0 ie Substituting equations (33) and (34) into (29) and (30), re- C(r,r )= C0 exp A(ξ) · dξ . (42) ~ Γ 0 spectively, and using (28), we have sl r Substituting last equation into (28) and using the initial condi- h i mω ∂ 1 0 0 tion (10), we readly obtain C0 = . Therefore the complete i~ + eA j(r)+ eFjk(xk−xk) C(r,r ) = 0, (35) 2πi~ ∂x j 2 Feynman propagator for a charged particle under the influence h i of a constant and uniform magnetic field takes the form ∂ 0 1 0 0 i~ − eA j(r )+ eFjk(xk−x ) C(r,r ) = 0, (36) 0 k 0 ∂x j 2 K(x,x ;τ) r ½ Z ¾ mω m ie r where we defined Fjk = ε jk3 B. = exp A(ξ) · dξ Our strategy to solve the above system of differential equa- 2πi~ sin(ωτ) 2πi~τ ~ r0 ½ ¾ ½ 0 2 ¾ tions is the following: we first equation (35) assuming in this im im (x3 − x ) 0 exp cot(ωτ)(r − r 0)2 exp 3 , (43) equation variables r as constants. Then, we impose that the 2~ 2~ τ result thus obtained is a solution of equation (36). With this goal, we multiply both sides of (35) by dx j and sum over j, to where in the above equation we omitted the symbol Γsl but, of obtain course, it is implicit that the line integral must be done along µ ¶ · ¸ a straightline, and we brought back the free propagation along ¡ ¢ 1 ∂C ie 1 0 the OX 3 direction. A few comments about the above result dx j = A j(r) + Fjk xk − xk dx j . (37) C ∂x j ~ 2 are in order.

Integration of the previous equation leads to 1. Firstly, we should emphasize that the line integral which appears in the first exponencial on the r.h.s. of (43) R 0 0 0 0 { ie r [A (ξ)+ 1 F ξ −x0 ] dξ } must be evaluated along a straight line between r and C(r,r ) = C(r ,r ) e ~ Γ r 0 j 2 jk( k k) j , (38) r. If for some reason we want to choose another R r where the line integral is assumed to be along curve Γ, to be path, instead of integral r0 A(ξ) · dξ, we must evaluate R r 0 specified in a moment. As we shall see, this line integral does r0 [A(ξ) − (1/2)B × (ξ − r )] · dξ. 0 not depend on the curve Γ joining r and r, as expected, since 2. Since we solved the Heisenberg equations for the gauge the l.h.s. of (37) is an exact differencial. 0 0 invariant operators R⊥ and π⊥, our final result is written In order to determine the differential equation for C(r ,r ) for a generic gauge. Note that the gauge-independent we must substitue expression (38) into equation (36). Doing and gauge-dependent parts of the propagator are clearly that and using carefully the fundamental theorem of differen- separated. The gauge fixing corresponds to choose tial calculus, it is straightforward to show that a particular expression for A(ξ). Besides, from (43) we imediately obtain the transformation law for the ∂C 0 0 propagator under a gauge transformation A → A + ∇Λ, 0 (r ,r ) = 0 , (39) ∂x j namely,

ie ie 0 0 0 Λ(r) 0 − Λ(r 0) which means that C(r ,r ) is a constant, C0, independent of K(r,r ;τ) 7−→ e ~ K(r,r ;τ)e ~ . r 0. Noting that Although this transformation law was obtained in a par- ¡ 0¢ ¡ 0 ¢ [B × ξ − r ] j = −Fjk ξk − xk , (40) ticular case, it can be shown that it is quite general. 1264 Brazilian Journal of Physics, vol. 37, no. 4, December, 2007

3. It is interesting to show how the energy spectrum (Lan- frequency ω0. Using the same notation as before, we can write dau levels), with the corresponding degeneracy per unit the hamiltonian of the system in the form area, can be extracted from propagator (43). With this purpose, we recall that the partition function can P2 H = H + 3 , (46) be obtained from the Feynman propagator by taking ⊥ 2m τ = −i~β, with β = 1/(K T), and taking the spatial B where trace, 2 2 Z ∞ Z ∞ π1 + π2 1 2 ¡ 2 2¢ H⊥ = + mω0 X1 + X2 . (47) Z(β) = dx1 dx2 K(r,r;−i~β) . 2m 2 −∞ −∞ As before, the Feynman propagator for this problem Substituting (43) into last expression, we get 0 0 (0) 0 takes the form K(x,x ;τ) = K⊥(r,r ;τ)K3 (x3,x3;τ), with Z ∞ Z ∞ (0) 0 mω K3 (x3,x3;τ) given by equation (14). The propagator in the Z(β) = dx1 dx2 , −∞ −∞ 2π~senh(~βω) OX 1X2-plane satisfies the differential equation (15) and will be determined by the same used in the previous example. where we used the fact that sin(−iθ) = −i sinh θ. Ob- Using hamiltonian (47) and the usual commutation rela- serve that the above result is divergent, since the area of tions the Heisenberg equations are given by the OX X plane is infinite. This is a consequence of 1 2 dR(τ) Π(τ) the fact that each Landau level is infinitely degenerated, = , (48) though the degeneracy per unit area is finite. In order dτ m dΠ(τ) to proceed, let us assume an area as big as we want, but = 2ωCΠ(τ) − mω2R(τ) , (49) finite. Adopting this kind or regularization, we write dτ 0 Z Z where we have used the matrix notation introduced in (17) L/2 L/2 L2 mω dx1 dx2 K (r,r;−i~β) ≈ and (20). Equation (48) is the same as (18), but equation (49) −L/2 −L/2 2π~senh(~βω) contains an extra term when compared to (19). In order to L2 eB decouple equations (48) and (49), we differentiate (48) with = ³ ´ respect to τ and then use (49). This procedure leads to the ~βω −~βω 2π~ e − e following uncoupled equation

− 1 ~βω d2R(τ) dR(τ) L2 eB e 2 c − 2ωC + ω2R(τ) = 0 (50) = ³ ´ dτ2 dτ 0 2π~ 1 − e−~βωc After solving this equation, R(τ) and Π(τ) are constrained to ∞ 2 L eB −β(n+ 1 )~ω satisfy equations (48) and (49), respectively. A straightfor- = ∑ e 2 c , ward algebra yields the solution n=0 2π~ R(τ) = M−R(0) + NΠ(0) (51) where we denoted by ωc = eB/2m the ciclotron fre- + 2 2 quency. Comparing this result with that of a parti- Π(τ) = M Π(0) − m ω0NR(0) , (52) tion function whose energy level E has degeneracy g , n n where we defined the matrices given by sin(Ωτ) N = eωτC (53) Z(β) = g e−βEn , ∑ n mΩh i n ω M± = eωτC cos(Ωτ)1l ± sin(Ωτ)C , (54) Ω we imediately identify the so called Landau leves and q the corresponding degeneracy per unit area, 2 2 and frequency Ω = ω + ω0. Using (51) and (52), we write µ ¶ 1 g eB Π(0) and Π(τ) in terms of R(τ) and R(0), E = n + ~ω ; n = (n = 0,1,...) . n 2 c A 2π~ Π(0) = N−1R(τ) − N−1M−R(0), (55) h i + −1 + −1 − 2 2 Π(τ) = M N R(τ)− M N M +m ω0N R(0). (56) B. Charged harmonic oscillator in a uniform magnetic field Now, we must order appropriately the hamiltonian operator T 2 T In this section we consider a particle with mass m and H⊥ = Π (0)Π(0)/(2m)+mω0R (0)R(0)/2, which, with the charge e in the presence of a constant and uniform magnetic aid of equation (55), can be written as field B = Be and submitted to a 2-dimensional isotropic har- h i 3 T −1 T T − T −1 T monic oscillator potential in the OX 1X2 plane, with natural H⊥ = R (τ)(N ) − R (0)(M ) (N ) A. Aragao˜ et al. 1265 h i −1 −1 − 2 T 0 × N R(τ) − N M R(0) + mω0R (0)R(0) into (62), we write F(r,r ;τ) in the convenient form h i 2 h i mΩ2 mΩ 2 d cos(ωτ) = RT (τ) − RT (0)(M−)T F(r,r0;τ) = (r2 + r0 )csc(Ωτ)2+ mΩr · r0 2sin2 (Ωτ) 2 dτ sin(Ωτ) h i h i − 2 T d sin(ωτ) cos(Ωτ) × R(τ) − M R(0) + mω0R (0)R(0) + mΩr · Cr0 − i~Ω . (65) dτ sin(Ωτ) sin(Ωτ) 2 h mΩ T T − = 2 R (τ)R(τ) − R (τ)M R(0) Inserting this result into the differential equation 2sin (Ωτ) i −RT (0)(M−)T RT (τ) + RT (0)(M−)T M−R(0) ∂ i~ hr,τ|r0,0i = F(r,r0;τ)hr,τ|r0,0i, 2 T ∂τ +mω0R (0)R(0) and integrating in τ, we obtain 2 h mΩ 2 T − = R (τ) − R (τ)M R(0) 0 ½ h 2 C(r,r ) imΩ 2 2sin (Ωτ) i hr,τ|r0 , 0i = exp (r2 + r0 )cot(Ωτ) T − T 2 sin(Ωτ) 2~ −R (0)(M ) R(τ) + R (0) , (57) µ ¶ ¾ cos(ωτ) sin(ωτ) i − 2 r · r0 + r · Cr0 . (66) where superscript T means transpose and we have used the sin(Ωτ) sin(Ωτ) properties of the matrices N and M− given by (53) and (54). 0 In order to get the right time ordering, observe first that where C(r,r ) is an arbitrary integration constantto be determined by conditions (29), (30) and (31). Using (56) we can h¡ ¢ i T − T T − − calculate the l.h.s. of condition (29), R (0)(M ) R(τ) = R (τ)M R(0) + M R(0) i ,Xi(τ) , mΩ n where hr, τ |π (τ)|r0,0i = cos(Ωτ)x − cos(ωτ)x0 j sin(Ωτ) j j h¡ ¢ i h i h i o − − T i~ ω 0 0 M R(0) ,Xi(τ) = i~Tr N(M ) = sin(2Ωτ). + sin(Ωτ)x − sin(ωτ)x ε hr,τ|r ,0i, (67) i mΩ Ω k k jk3 Using the last two equations into (57) we rewrite the hamil- and using (55) we get the l.h.s. of condition (30), tonian in the desired ordered form, namely, n 0 mΩ 0 2 h hr, τ |π j(0)|r ,0i = cos(ωτ)x j − cos(Ωτ)x j mΩ 2 2 T − sin(Ωτ) H⊥ = 2 R (τ) + R (0) − 2R (τ)M R(0) h i o 2sin (Ωτ) ω 0 0 + sin(Ωτ)x − sin(ωτ)xk ε jk3 hr,τ|r ,0i. (68) i~ i Ω k − sin(2Ωτ) . (58) mΩ With the help of the simple identities For future convenience, let us define ∂ 2 02 ∂ 2 02 0 ω (r + r ) = 2x j ; 0 (r + r ) = 2x j U(τ) = cos(ωτ)cos(Ωτ) + sin(ωτ)sin(Ωτ), (59) ∂x j ∂x j Ω ω ∂ 0 0 ∂ 0 V(τ) = sin(ωτ)cos(Ωτ) − cos(ωτ)sin(Ωτ) (60) r · r = x j ; 0 r · r = x j Ω ∂x j ∂x j ∂ ∂ and write matrix M−, defined in (54), in the form 0 0 0 r · Cr = ε jk3xk ; 0 r · Cr = −ε jk3xk. ∂x j ∂x j M− = U(τ)1l + V(τ)C. (61) and also using equation (66), we are able to compute the right Substituting (61) in (58) we have hand sides of conditions (29) and (30), which are given, respectively, by 2 h mΩ 2 2 T H⊥ = R (τ) + R (0) − 2U(τ)R (τ)R(0) n 0 2sin2 (Ωτ) i~ ∂C(r,r ) cos(Ωτ) cos(ωτ) 0 − 0 + mΩ x j − mΩ x j i C(r,r ) ∂x j sin(Ωτ) sin(Ωτ) T i~ − 2V(τ)R (τ)CR(0) − sin(2Ωτ) . (62) sin(ωτ) o mΩ − mΩ ε x0 − eA (r) hr,τ|r0,0i (69) sin(Ωτ) jk3 k j The next step is to compute the classical function F(r,r0;τ). Using the following identities and h i n 0 ΩU(τ) d cos(ωτ) i~ ∂C(r,r ) cos(Ωτ) 0 cos(ωτ) 2 = − , (63) 0 0 mΩ x j + mΩ x j sin (Ωτ) dτ sin(Ωτ) C(r,r ) ∂x j sin(Ωτ) sin(Ωτ) h i ΩV(τ) d sin(ωτ) sin(ωτ) o = − , (64) − mΩ ε x − eA (r0) hr,τ|r0,0i. (70) sin2 (Ωτ) dτ sin(Ωτ) sin(Ωτ) jk3 k j 1266 Brazilian Journal of Physics, vol. 37, no. 4, December, 2007

Equating (67) and (69), and also (68) and (70)), we get the Note that the integrand has a vanishing curl so that we can system of differential equations for C(r,r0) choose the path of integration Γ at our will. Choosing, as before, the straight line between r0 and r, it can be shown that 0 h i ∂C(r,r ) Fjk 0 i~ + e A j(r) + xk C(r,r ) = 0, (71) ∂x 2 Z r h i Z r j B 1 0 0 h i A(ξ) − × ξ · dξ = A(ξ) · dξ + Br · Cr , (75) ∂C(r,r ) 0 Fjk 0 0 r0 2 r0 2 i~ 0 − e A j(r ) + xk C(r,r ) = 0. (72) ∂x j 2 where, for simplicity of notation, we omitted the symbol Γsl Proceeding as in the previous example, we first integrate (71). indicating that the line integral must be done along a straight With this goal, we multiply it by dx j, sum in j and integrate it line. From equations (73), (74) e (75), we get to obtain ½ ¾ ½ Z ¾ ½ ¾ Z r h i r ie Fjk 0 ie im 0 0 0 0 C(r,r ) = C0 exp A(ξ) · dξ exp r · Cr , (76) C(r,r ) = C(r ,r )exp A j(ξ) + ξk dξ j , 0 ~ Γ r0 2 ~ r ~ (73) where the path of integration Γ will be specified in a moment. which substituted back into equation (66) yields Inserting expression (73) into the second differential equation ½ Z ¾ (72), we get C ie r hr,τ|r0,0i = 0 exp A(ξ) · dξ ∂ sin(Ωτ) ~ r0 C(r 0,r 0) = 0 =⇒ C(r 0,r 0) = C , n n 0 0 imΩ 2 ∂x j exp (r2 + r0 )cos(Ωτ) 2~sin(Ωτ) where C is a constant independent of r 0, so that equation h ω i oo 0 −2r · r0 cos(ωτ) − 2 sin(ωτ) − sinΩτ rCr0 (77) (73) can be cast, after some convenient rearrangements, into Ω the form ½ ¾ Z r h i The initial condition implies C0 = mΩ/(2πi~). Hence, the 0 ie 1 C(r,r ) = C0 exp A(ξ) − B × ξ · dξ . (74) desired Feynman propagator is finally given by ~ Γ r0 2

0 0 (0) 0 K(x,x ;τ) = K⊥(r,r ;τ)K3 (x3,x3;τ) r ½ Z ¾ ½ n mΩ m ie r imΩ 2 = exp A(ξ) · dξ exp cos(Ωτ)(r2 + r0 ) 2πi~ sin(Ωτ) 2πi~τ ~ r0 2~sin(Ωτ) ½ ¾ h ω i oo im (x − x0 )2 − 2cos(ωτ)r · r0 − 2 sin(ωτ) − sin(Ωτ) r · Cr0 exp 3 3 , (78) Ω 2~ τ

where we brought back the free part of the propagator cor- same comments done before are still valid here, namely, the responding to the movement along the OX 3 direction. Of above expression is written for a generic gauge, the transfor- course, for ω0 = 0 we reobtain the propagator found in our mation law for the propagator under a gauge transformation is first example and for B = 0 we reobtain the propagator for a the same as before, etc. We finish this section, extracting from bidimensional oscillator in the OX 1X2 plane multiplied by a the previous propagator, the corresponding energy spectrum. free propagator in the OX 3 direction, as can be easily checked. With this purpose, we first compute the trace of the propaga- Regarding the gauge dependence of the propagator, the tor,

½ ¾ Z ∞ Z ∞ Z ∞ Z ∞ h ³ ´ i 0 mΩ imΩ 2 2 dx1 dx2 K⊥(x1,x1,x2,x2;τ) = dx1 dx2 exp 2 cos(Ωτ) − cos(ωτ) (x1 + x2) −∞ −∞ 2πi~ sin(Ωτ) −∞ −∞ 2~ sin(Ωτ) 1 = , (79) 2[cos(Ωτ) − cos(ωτ)]

where we used the well known result for the Fresnel integral. Using now the identity cos(Ωτ) − cos(ωτ) = −2 sin[(Ω + ω)τ/2] sin[(Ω − ω)τ/2)], A. Aragao˜ et al. 1267 we get for the corresponding energy Green function constant and uniform magnetic field (Landau problem) and the same problem in which we added a bidimensional har- Z ∞ Z ∞ Z ∞ i Eτ 0 monic oscillator potential. Although these problems had al- G (E) =−i dτe ~ dx1 dx2 K⊥(x1,x1,x2,x2;τ) 0 −∞ −∞ ready been treated from the point of view of Schwinger’s ac- Z i Eτ i ∞ e ~ tion principle, the novelty of our work relies on the fact that = dτ Ω+ω Ω−ω we solved the Heisenberg equations for gauge invariant opera- 4 0 sen( τ)sen( τ) Ã2 2 !Ã ! tors. This procedure has some nice properties, as for instance: Z ∞ ∞ ∞ i Eτ −(l+ 1 )(Ω+ω)τ −i(n+ 1 )(Ω−ω)τ (i) the Feynman propagator is obtained in a generic gauge; = −i dτe ~ ∑ e 2 ∑ e 2 , 0 l=0 n=0 (ii) the gauge-dependent and gauge-independent parts of the propagator appear clearly separated and (iii) the transforma- where is tacitly assumed that E → E − iε and we also used tion law for the propagator under gauge transformation can that (with the assumption ν → ν − iε) be readly obtained. Besides, we adopted a matrix notation which can be straightforwardly generalized to cases of rel- ∞ 1 −i(n+ 1 )ντ ativistic charged particles in the presence of constant elec- = 2i e 2 . ν ∑ tromagnetic fields and a plane wave electromagnetic field, sen( 2 τ) n−0 treated by Schwinger [2]. For completeness, we showed ex- Changing the order of integration and summations, and inte- plicitly how one can obtain the energy spectrum directly from grating in τ, we finally obtain que Feynman propagator. In the Landau problem, we obtained the (infinitely degenerated) Landau levels with the cor- ∞ 1 responding degeneracy per unit area. For the case where we G(E) = , (80) ∑ included the bidimensional harmonic potential, we obtained l,n=0 E − Enl the energy spectrum after identifying the poles of the corre- where the poles of G(E), which give the desired energy levels, sponding energy Green function. We hope that this pedagogi- are identified as cal paper may be useful for undergraduate as well as graduate students and that these two simple examples may enlarge the Enl = (l + n + 1)~Ω + (l − n)~ω, (l,n = 0,1,...) . (81) (up to now) small list of non-relativistic problems that have been treated by such a powerful and elegant method. The Landau levels can be reobtained from the previous result by simply taking the limit ω0 → 0: 1 E −→ (2l + 1)~ω = (l + )~ω , (82) nl 2 c with l = 0,1,... and ωc = eB/m, in agreement to the result we had already obtained before. Acknowledgments

F.A. Barone, H. Boschi-Filho and C. Farina would like to IV. FINAL REMARKS thank Professor Marvin Goldberger for a private communication and for kindly sending his lecture notes on quantum In this paper we reconsidered, in the context of Schwinger’s mechanics where this method was explicitly used. We would method, the Feynman propagators of two well known prob- like to thank CNPq and Fapesp (brazilian agencies) for partial lems, namely, a charged particle under the inﬂuence of a ﬁnancial support.

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